Find the range of values of x that satisfy the inequality (x+1)(x-2)

Question

Find the range of values of x that satisfy the inequality \(\frac{(x+1)(x-2)}{(x-5)(x+3)} > 0\)

A. x < -3
B. -1 < x < 2
C. x < -3 or -1 < x < 2 or x > 5
D. -1 < x < 2 or x > 5
E. x < -3 or -1 < x < 2

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Wavy Line Method Application has been explained in detail in the following post:: https://gmatclub.com/forum/wavy-line-method-application-complex-algebraic-inequalities-224319.html

Detailed solution will be posted soon.

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Official Answer

C

EgmatQuantExpert · Accepted Answer

Solution  :

Hey Everyone,

Please find below the solution of the given problem,

Plotting the zero points and drawing the wavy line:

https://e-gmat.com/blogs/wp-content/uploads/2016/11/S4.jpg

Required Range  :

x < -3 or -1 < x < 2 or x > 5

Correct Answer  : Option C

https://e-gmat.com/blogs/wp-content/uploads/2018/08/Ability-Quiz-Algebra-e1533292163516.png

abhimahna · Answer

Equation is  \frac{(x+1)(x-2)}{(x-5)(x+3)} > 0

We have the values of x as -3,-1,2 and 5.

Since x cannot be equal to 5 or -3 we need to exclude those values(they are at the denominator and we cannot have 0 at the denominator)

So, substituting these values of the number line, we will the range of x as

(-infinity,-3) U (-1,2) U (5,infinity)

Please correct me if I am missing anything.

gmat800live · Answer

VeritasKarishma  apologies for bothering you. I read your inequality series and it's been enlightening.

I was wondering, for a problem such as this one where you have (x+1)(x-2) on numerator and (x-5)(x+3) on denominator... how would you approach it? Clearly the "wavy" method works... but I wonder how would you tackle this with the method you explain in your series "Inequalities with complications".

Am I allowed to multiply by 1 since it's positive? For instance I multiply (x-5)/(x-5) and (x+3)/(x+3) on left side to get (x+1)(x-2)(x-5)(x+3) on numerator and then (x-5)^2 and (x+3)^2 on denominator! This means that I can ignore the denominator since it will be always positive, and hence I end with a structure similar to the ones you tackled on your post (x+1)(x-2)(x-5)(x+3) > 0.

Would that work? Thank you.

KarishmaB · Answer

You don't need to make it so complicated though what you suggest works too. The reason why factors in the denominator work is simply this:

When is abcd positive? When a, b, c and d all are positive. Or when a and b are positive and c and d are negative. etc etc

When is ab/cd positive? When a, b, c and d all are positive. Or when a and b are positive and c and d are negative. etc etc

Aren't the two cases the same? (Except that c and d cannot be 0 in the second case) Anyway, since the expression needs to be positive, a, b, c and d cannot be 0 in first case either.

Does it matter whether c and d are in numerator or denominator? No. 
The signs of a, b, c and d decide the sign of the expression irrespective of whether they are in numerator or denominator. The case is the same here.

uttam94317 · Answer

We have the values of x as -3,-1,2 and 5.

So, substituting these values of the number line, we will the range of x as

kunalbean · Answer

Hello! Can you please explain how did you get x < -3? The eqn = x -5 > 0 i.e. x >5. So how come for x+3 > 0 it becomes x<-3? Shouldnt it be x>-3?

Stanindaw · Answer

Where can i get the wavy ine method?

KarishmaB · Answer

Check this post: https://gmatclub.com/forum/inequalities ... 91482.html

vrinda6 · Answer

Hey , can you please explain how did you get:

-1 < x < 2 ?

As per the numerator, and the avy method and if i were to employ that on the number line- I should get:

x>2 for positive values
-1<x<2 for negative value 
x<-1 for positive values

Bunuel · Answer

Find the range of values of x that satisfy the inequality \(\frac{(x+1)(x-2)}{(x-5)(x+3)} > 0\)

A. x < -3
B. -1 < x < 2
C. x < -3 or -1 < x < 2 or x > 5
D. -1 < x < 2 or x > 5
E. x < -3 or -1 < x < 2

We have:

\(\frac{(x+1)(x-2)}{(x-5)(x+3)} > 0\)

The transition points, in ascending order, are \(x = -3\), \(x = -1\), \(x = 2\), and \(x = 5\) (these are the values of x at which the expression changes sign). This gives us five ranges:

\(x < -3\)
\(-3 < x < -1\)
\(-1 < x < 2\)
\(2 < x < 5\)
\(x > 5\)

Next, test an extreme value for x: if x is a large enough number, say 100, then both the numerator and denominator will be positive, resulting in a positive value for the whole expression. Therefore, when \(x > 5\), the expression is positive.

Now, here’s the trick: since the expression is positive in the 5th range, it will be negative in the 4th range, positive again in the 3rd range, negative again in the 2nd range, and positive again in the 1st range, following the pattern: +, -, +, -, +.

Thus, the expression is positive for \(x < -3\), \(-1 < x < 2\), and \(x > 5\).

Answer: C.

For more check inequalities chapter here: https://gmatclub.com/forum/ultimate-gma ... equalities

vrinda6 · Answer

Thankyou for the detailed explanation. However just want to confirm, if we have a situation like this when we have multiple critical points do we plot all of them together to find the postive/negative zones?

Bunuel · Answer

For more check inequalities chapter here: https://gmatclub.com/forum/ultimate-gma ... equalities

Find the range of values of x that satisfy the inequality (x+1)(x-2)

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Find the range of values of x that satisfy the inequality (x+1)(x-2)

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

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